A partical moves at a speed of 3.0 m/s in the + x-direction. Upon reachinng the origin, the partical receives a continuous constant acceleration of 0.75 m/s^2 in the -y direction. what is the position of the partical in 4.0 s later?
2007-02-24 16:45:00 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
so the position vector relative to the origin is
r=3t i - 1/2*.75 t^2 j
in 4 seconds, the position vector is 12i-6j. Taking the magnitude we get |r|=sqrt(12^2+(-6)^2)=sqrt(144+36)=13.42 m
2007-02-24 16:55:36 · answer #1 · answered by Rob M 4 · 0⤊ 1⤋
The x-coordinate will just be:
3.0m/s * 4.0s = 12 m
The y-coordinate is calculated using constant acceleration equations:
y = (Vi)t + .5at^2
where Vi (initial velocity IN THE Y DIRECTION) is zero.
y = 0t + .5 (-.75)(4.0)^2
y = -6 m
the coordinates are (12, -6)
2007-02-25 00:52:21 · answer #2 · answered by J 2 · 0⤊ 1⤋